If the angle of dip at places $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively,the ratio of the horizontal component of the Earth's magnetic field at $A$ to that at $B$ will be.
$[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 30^{\circ}=\frac{1}{2}, \quad \cos 30^{\circ}=\frac{\sqrt{3}}{2}]$

  • A
    $\sqrt{2}: 1$
  • B
    $1: \sqrt{2}$
  • C
    $\sqrt{2}: \sqrt{3}$
  • D
    $\sqrt{3}: \sqrt{2}$

Explore More

Similar Questions

The Earth's magnetic field was flipped by $180^{\circ}$ a million years ago. This flip was relatively rapid and took $10^5 \, \text{years}$. Then, the average change in orientation per year during the flip was closest to ............ $s$.

The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be:

The magnetic field of the Earth at the equator is approximately $4 \times 10^{-5} \, T$. The radius of the Earth is $6.4 \times 10^6 \, m$. Then the dipole moment of the Earth will be nearly of the order of:

The angle of dip at a place is $30^{\circ}$,and the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, T$. Calculate the vertical component and the total magnetic field at that place.

The horizontal component of Earth's magnetic field at a place is $3.5 \times 10^{-5} \,T$. $A$ very long straight conductor carrying a current of $\sqrt{2} \,A$ in the direction from South-East to North-West is placed. The force per unit length experienced by the conductor is $..............$ $\times 10^{-6} \,N/m$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo